Form a quadratic equation whose one root is 3 | vedclass

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Question

(A) Given that the sum of the roots is $6$.One root is $3-\sqrt{5}$.Let the roots be $\alpha$ and $\beta$. Then $\alpha = 3-\sqrt{5}$ and $\alpha + \b

Vedclass · Accepted Answer

$x^{2}-6x+4=0$

Vedclass · Answer

(A) Given that the sum of the roots is $6$.One root is $3-\sqrt{5}$.Let the roots be $\alpha$ and $\beta$. Then $\alpha = 3-\sqrt{5}$ and $\alpha + \beta = 6$.Substituting the value of $\alpha$,we get $(3-\sqrt{5}) + \beta = 6$.Therefore,$\beta = 6 - (3-\sqrt{5}) = 3+\sqrt{5}$.The product of the roots is $\alpha \cdot \beta = (3-\sqrt{5})(3+\sqrt{5})$.Using the identity $(a-b)(a+b) = a^2 - b^2$,we get $3^2 - (\sqrt{5})^2 = 9 - 5 = 4$.$A$ quadratic equation is given by $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.Substituting the values,we get $x^2 - 6x + 4 = 0$.

Form a quadratic equation whose one root is $3-\sqrt{5}$ and the sum of roots is $6$.

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